/ 


I 


GROUPS  OF  SPECIAL  GALOIS 
DOMAINS  OF  RATIONALITY 


BY 


HARRY  ALBERT  BENDER 
A.  B.  Ohio  University,  1918 


THESIS 


Submitted  in  Partial  Fulfillment  of  the  Requirements  for  the 


Degree  of 


MASTER  OF  ARTS 
IN  MATHEMATICS 


IN 


THE  GRADUATE  SCHOOL 


OF  THE 


UNIVERSITY  OF  ILLINOIS 


1921 


Digitized  by  the  Internet  Archive 
in  2015 


https://archive.org/details/groupsofspecialgOObend 


UNIVERSITY  OF  ILLINOIS 


THE  GRADUATE  SCHOOL 


I HEREBY  RECOMMEND  THAT  THE  THESIS  PREPARED  UNDER  MY 


BE  ACCEPTED  AS  FULFILLING  THIS  PART  OF  THE  REQUIREMENTS  FOR 


THE  DEGREE  OF 


/3 \CH&f/&L. 


In  Charge  of  Thesis 
'/  Head  of  Department 


Recommendation  concurred  in* * 


Committee 

on 

Final  Examination* 


*Required  for  doctor’s  degree  but  not  for  master’s 


Let  f(x)=0  be  an  irreducible  equation  of  degree  n and  let 

x,,x^  ,Xj  , ,x^  be  its  roots.  Let  p*  be  sin  arbitrary  rational 

prime  and  let  us  suppose  that  in  k(p) 

(l)  f (x)=f,  (x)  .f2  (x)  .fj  (x)  fs-(x)  (p) 

where  fc-(x)  is  of  degree  n^  ,and  irreducible  in  k(p)  . 

In  k(p),the  domain  of  the  rational  p-adic  numbers,the 
number  x,  cannot  satisfy  an  equation  of  degree  less  than  n (x,  is 
taken  in  this  sense  as  any  one  of  the  roots  of  f(x)).  For  suppose 
that 

0 ( x ) =a0  x*'4a;  x** 

is  a polynomial  of  degree  less  than  n.with  p-adic  coefficients, 
such  that  0( x; )=0  (p) ,and  let 

X/'-a^w, -h a^w.,* + a^w*  ( i=1 ,2,3, ,n)  . 

We  then  have 

<P(*f  =A,  w, -f  -+A^w^ 

Ai=afl^'+a/  + + \.fa/c  (i=1  >2,3, ,n)  . 

Since  an  integer  of  k(x,)  is  divisible  by  p when  and  only  when 
each  coefficient  in  its  representation  by  a fundamental  system 
is  a multiple  of  p,we  conclude  that  <^(xy)=0  (p)  when  and  only 

when  A^=0  (p)  (i=1,2,3, --,n),and  from  this  system  of 

equations  it  follows  that  Ja t- J.a^=0  (p)  . 

But  the  a(^are  ordinary  rational  integers  and  hence  |aiy.'| 
is  a number  of  k ( 1 ) and  since  it  is  not  zero  there  exists  a 

*By  p without  subscripts  we  shall  mean  an  arbitrary  rational 
prime, and  p with  subscripts  we  shall  mean  the  prime  divisors 

of  p in  the  algebraic  domain. 


. 


/4 yM~a4  =,a-4-°  (p)- 

Hence  ^(x)  must  vanish  Identically , and  x,  cannot  in  k(p)  satisfy 
an  equation  of  degree  less  than  n. 

From  ( 1 ) we  have 

f ( x,)  =f,(x,)  •fJL(xl)  .f3[x,)  . fj( Xj)  =0  ( p ) 

while  no  one  of  the  factors  is  zero.  We  therefore  conclude  that 
in  this  case  B(p,x,)  is  not  a domain. 

Since  f(x)  is  irreducible  in  the  ordinary  sense  its 
discriminant  cannot  vanish  and  hence  it  must  also  be  different 
from  zero  for  the  domain  of  p.  Consequently  no  two  of  the  £ 
factors  can  be  equal. 

We  shall  now  introduce , corresponding  to  each  of  the  s 
factors  of  f(x)  ,s  new  systems  of  values  for  the  numbers  of  R(p,X/) 
as  follows.  If/?=B(x/)  is  any  number  of  R(p,x/)  and  if 

B(x)=Qt(x)  .fc-  (x)  + R<  (x)  (p), 

we  shall  call  R i(xf)  the  value  of  (3  for  the  domain  of  p-  cor- 
responding to  the  factor  f^(x)  and  shall  write /3=Rc (x,  ) (p- ) . 

Two  numbers  f3,- B/(x/  ) and/^=B^(x^  ) are  said  to  be  equal 
for  the  domain  p^  when  and  only  when  B/(x)-B^(x)  is  divisible 
by  U (x) . 

We  thus  have  s new  rings  R(p^  ,x,  ) (i= 1,2,3* -,s)  such 

that  each  number  of  R(p,x,)  is  for  the  domain  of  p^-  equal  to 
some  number  of  R(jj£  ,x/  ) and  the  sum , difference , and  product  of 
two  numbers  of  R(p,x^  ) is  for  the  domain  of  equal  to  the  sum, 
difference , and  product  respectively  of  the  corresponding  numbers 
of  R(p^  ,x/  ) . Evidently  % (x,  )=.0  (p^  ) 


. . « £JH 

■ 


5 


We  shall  next  see  that  these  p -adlc  values  of  the  numbere 
of  R(p,Xy)  constitute  a domain.  We  need  only  show  that  every 
number/3  ^0  (p^  ) has  a uniquely  determined  reciprocal  in  R(px-,x;). 

Let  us  therefore  suppose  that  ft  =B(xy ) rjfcO  Since/?  £0  (p^-) 

it  follows  that  B(x)  is  not  divisible  by  f/  (x)  and  hence  since 
fy  (x)  is  irreducible  we  know  that  they  are  relative  prime.  Hence 
there  are  two  polynomials  \f^(x)  and  such  that 

(p^x)  .fy  (x)  -fl£(x)  .B(x)  = 1 (p) 


and  since  rational  numbers  are  equal  for  the  domain  of  p - ,we 
can  write 

(p/x)  .fy  (x)-/-(^(x)  .B(x)  = 1 (py  ) 

the  coefficients  of  the  polynomials  being  rational  numbers.  But 

f/ (xf  )=0  (p^ ) 

and  hence  (x, ) ).  Therefore/?  has  a reciprocal  which  is 

unique, for  if  q and  /?,  are  two  numbers  such  that/?  ./>  -n,A- 1 (p-  ) , 

j / y A. 

then  /0/-  ^=0  (py  ) and  hence  /?,/?( - /^)  -fi,  -fy= 0 (Py ) • 

Therefore  /?  = /?  (p.)  and  the  p -adic  values  of  the  numbers  of 
R(p,x/ ) form  a domain  which  we  shall  denote  by  k(py ,x( ) . 

In  k(x^)  p is  divisible  by  s distinct  prime  divisors 

%'/  'tyx  » ,v:here  the  notation  is  so  chosen  that 

xy ) =C^-j.  Since  in  k(p)  xy  cannot  satisfy  an  equation  of  degree 

less  than  n and  since  t: (x)  (j=1,2,3, -,s)  are  all  diBtinct, 

) (^rj)  • But  fj  (xy  ) .f^  (xy  ) . -f5(x^)=0  (p)  and 

hence  f,  (x^  ) .f^Uy  ) -- .f^  (xy  )=°  (jc-^  ) (i=1,2,— ,n)  (j=1»2,-s) 

( k=  1 , 2 , — , s ) • 

Let  us  consider  the  array  J constructed  with  the  various 
distinct  prime  divisors  of  p in  each  of  the  n domains  k(xy) 


' 


. 


4 


(i=1,2,3» ,n)  in  which  each  row  contains  the  distinct  prime 

divisors  of  p in  a certain  one  of  the  domains. 


p// 

■P/JL  ’ 

p,J  • 

1 

1 

1 

1 

m 

L 

,vxx  * 

p«  • 

™*P« 

p*, 

P*J  • 

pyy  *" 

*9h  s 

and  associated  with  this  the  array  of  ns  numbers  from  the  domains 

fy  ( X/  ) > f 2 ( Xy  ) , f J ( X/  ) > , f^(  X,  ) 

f / ( ) > fz  ( ^2.  ^ ) i » f j-  ( ) 

n 

f / ^ ) > f ) j fj  ( X^  ; f — t f j-  ( X*  ) • 

We  observe  that  this  arrangement  has  been  so  made  that  any 
element  in  II  is  zero  for  the  domain  of  the  corresponding  element 
of  I. 

Let  us  next  suppose  that  F(V)=0  is  the  Galoisian  resolvent 
of  f(x)=0  for  the  domain  R(l),and  that  t is  any  one  of  its  roots, 
It's  degree  we  shall  suppose  to  be  g and  when  we  need  to  dis- 
tinguish between  the  roots  we  shall  denote  them  by  V;  , V3L,----,V^. 
The  substitutions  on  the  n roots  by  which  these  V's  are  derived 
from  V/  form  a transitive  group  G of  order  g.  The  distinct  prime 

divisors  of  p in  k(V)  we  shall  denote  by  P , , ?3  , , . 

Since  k(x, ) is  included  in  k(v)  we  see  that  t^s . 

Let  us  suppose  that  in  k(p) 

F(V)=F,  (V)  -Fa(v)  -Fj  (V)  F*  (V)  (pj 

where  F,;  (V)  is  of  degree^'and  F^  (V)=0  (P^- ) ( i=1  ,2,3, ,t)  • 

And  since  the  factors  are  all  irreducible, they  are  all  distinct 
and  only  one  of  them  is  zero  for  the  domain  of  (P^  ) according  to 


. 


5 

the  same  discussion  as  in  the  case  of  f(x)=f/  (x)  .fa(x)  . .f^  (x)(pj 

Hence  each  is  a root  of  some  F^(v)=0  (P^)  and  since  this 
equation  cannot  have  more  roots  than  its  degree  we  know  that  at 

most  of  the  numbers  V#  , , , are  roots  of  F^(V)=0 

where  is  the  degree  of  f^(V).  Since  this  is  true  for  all 

(i=1,2, ,g)  and  for  all  F^{V)  (k=1,2, ,t)  and  since  H(H---+flpg 

we  know  that  each  of  the  equations  F^(v)=0  (PA)  (k=1,2, ,t) 

has  in  the  domain  k(PA  ,V)  exactly  roots. 

Let  S be  a substitution  of  G which  transforms  V.  into 
and  let  us  suppose  that  F^(V,  )=0  (PA).  The  substitution  S then 
transforms  F^CV,  ) into  ^(V^)  and  this  is  again  zero  for  the 
domain  of  some  P^.  If  F^V^  )=0  (P^)  th en/^=\.  In  this  case  we 
shall  say  that  S transforms  the  prime  divisor  P^  into  the  prime 
divisor  PA,or  the  substitution  S leaves  p*  invariant.  If  however 
F^(V t )£0  (P^‘  but  F^(V^-  )=0  (i^J^^^we  shall  say  that  S trans- 
forms the  prime  divisor  PA  into  the  prime  divisor 

Hence  every  substitution  of  G either  leaves  P^  invariant 
or  transforms  it  into  another  prime  divisor  of  p in  k(V) • 

If.  we  now  consider  that 

F/  (V)  -F ^(V) Fa  (V)=(V-fy  )(V-^) (V-.^)(^  ) 

and  factorization  of  a polynomial  in  a domain  is  unique  we  know 
that 

F;  ( V)  = ( V- V„  ) ( V-V/JL)  ( V- V/3 ) ( V-V^)  ( PA  ) 

FA  ( V)  = ( V- Vx, ) ( V-Vxx)  ( V- V^)  - (V-^)  ( P ) 

in  * 

E,  (V)  = { V-V„  ) ( V-T^)  (V-V^) (V-\r„)  {P*  ) . 

The  substitutions  of  G which  leave  P-,  invariant  form  a 


' 

• 

• 

6 

subgroup  of  G of  order  say  g^  . Let  us  call  this  subgroup  G/ . The 
substitutions  of  G/  form  a group, since  G contains  all  the  sub- 
stitutions which  when  operated  on  gives  all  the  other  V's,it 
then  contains  all  the  substitutions  which  when  operated  on  V^y 
gives  all  the  roots  of  F^(V)  in  and  by  hypothesis  all  these 
substitutions  are  in  G/  , thus  from  the  theorem  in  group  theory 
these  substitutions  of  G form  a group*. 

Every  substitution  of  G transforms  each  Vy  into  some  i£j 
unless  the  substitution  is  the  identical  substitution  because 
each  is  a primitive  number  of  k(V)  and  hence  under  any  sub- 
stitution S goes  over  into  one  of  its  conjugates  since  k(V)  is  a 
galois  field  or  domain. 

Let  us  suppose  that  H;^H^  (1=1,2, ,t)  and  1,  Sa  , , 

are  substitutioms  such  that  operating  on  V;  gives  V</(*  . Let  us 

represent  this  operation  as  IV^V,,  , , SjV/f=V/Jt , S^Vfl=V;iS. 

Thus  there  are  at  least  Hy  substitutions  that  leave  P^  invariant. 
Hence  the  order  of  Gy  is  <£Hy  . Moreover  P^  cannot  be  left  invariant 
by  any  other  substitution  of  G because  if  such  were  the  case  then 
this  substitution  S transforms  V/  into  and  since  and  S both 
transform  V/y  into  V,^  and  s/s  leaves  V7/  invariant.  But  this  is  as 
we  have  said  above, only  true  when  s/s  is  the  identical  substitu- 
tion or  when  S~.'s=  1 or  =S . Hence  the  order  of  Gy  is  H,  . 

G contains  at  least  t- 1 substitutions  which  do  not  leave 
invariant , namely  the  substitutions  which  carry  over  into  V^y 

(1=1,2,-— ,t)  or  tx =T£( , =VJ(, , Then  t,G, 

consists  of  Hy  substitutions  which  do  not  leave  invariant , for 

*Miller ,Blichf eldt ,and  Dickson  'Finite  Groups'  page  286. 


»• 


. 

- 


7 


since  Greaves  P^  invariant  it  will  give  a root  of  FjL(V)=0  (P^) 
when  operating  on  V^and  all  of  these  roots  are  distinct  from 
those  of  Fy  (V) , therefore  t^Gy  transforms  P^  into  another  prime 

divisor.  The  same  argument  holds  with  t3Gy  , t G^  , , trG/  . Thus 

G contains  at  least  Hy  substitutions  that  leave  P^  invariant , and 
(t-i)H  substitutions  that  will  transform  P^  into  some  other  prime 
divisor.  Therefore  G contains  at  least  tHy  substitutions , that  is 

S^tH/  . Since  g=H,-f  H3  + -fti;  and  Hyis^H*  (i=1,2,3, ,t)  we 

see  that  g^tHy  . We  therefore  conclude  that  g=tHy and  Hy  =H3L=HJ(==-=:H^ 

Hence  Fy  (V)  , F^(V)  , F^V), ,Fr(V)  are  all  of  the  same  degree, 

and  the  substitutions  of  Gy  permutes  the  V^y  in  TXT  with  the  same 
first  subscript  among  themselves. 

Since  G(V)  of  degree  nj ,and  whose  roots  are  derived  from 
the  nj  valued  function  of  Vy  is  reducible  in  k(p) ,and  Fy(V)  is 
that  irreducible  factor  in  k(p)  for  which  0 (P^),the  sub- 
stitutions on  Xy  , x^,  Xj  , — , x^  by  which  the  roots  of 

Fy (V)=0  (P^)  are  derived  from  Vf  is  called  the  group  of  the 
given  equation  in  k(p) . 

Since  k(Xy  ) is  included  in  k(V) 

fy  (x)  = (x-x// ) (x— x/JL)  (x-xyJ)-- (x-x^)=0  (P^) 

f^  (x)  = (x-xJ_/)  (x-x^Kx-Xjj) ( x-x^J)  =0  (P^  ) 

fj  (x)  - (x-x^)  (x-Xjj,)  (x-x^) (x-Xj^=0  (PA) 

where  xj^  ( j=1  ,2,3, ,s)  (i=1  ,2,3, ,n^  ) is  some  one  of  the 

roots  of  f ( x ) =0  • The  coefficients  of  f (x)=fy  (x)  *fj_(x)  «r--  .fj  (x) 

being  symmetric  functions  of  x^  (k=1,2, ,s)  (i=1,2, ,n  ) 

are  unaltered  numerically  by  the  substitutions  of  G,  , because  Gy 


8 


leaves  invariant , and  hence  equals  numbers  in  R.  It  has  been 
shown  that  Cr/  contains  all  the  substitutions  that  leave 
invariant . 

Since  the  roots  of  f^  (x)=0  (P^  ) are  x</?  x^  , x^  , , x^ 

and  the  roots  of  f^(x)  in  k(p)  are, say  y^  , , , y^, 

and  from  the  simple  isomorphism  that  exists  between  the  roots  in 
k(p)  and  those  in  P^  ,we  see  that  the  element  x--  in  the  substitu- 

<7 

tions  of  G,  can  be  replaced  by  y.-,and  we  have  the  group  of  f(x) 
in  k(p)  in  terms  of  the  roots  of  f(x)=0  in  k(p) . 

Thus  the  Theorem:  If  f(x)=0  is  irreducible  in  R( 1 ) 

and  its  group  in  R is  G,then  in  k(p)  the  group  of  f(x)=f  (x)  .fj_(x) 
fj  (x)  ‘fj  (x)  (p)  is  simple  isomorphic  to  a subgroup  G/  of  G. 

Since  this  group  is  simple  isomorphic  to  a subgroup  of  G, 
we  shall  discuss  some  of  the  properties  of  this  subgroup  instead 
of  the  group  in  k(p) • 

Since  f(x)  is  irreducible  in  R( 1 ) its  group  is  transitive 
in  the  n roots.  If  f(x)  is  irreducible  in  k(p)  its  group  will  be 
transitive  also.  Thus  in  this  case  is  G or  a transitive  sub- 
group of  G.  If  f ( x ) is  reducible  in  k(p)  its  group  is  intran- 
sitive in  the  n roots.  Since  (x)  , f^(x),  f^  (x)  , •,  fs  (x) 

have  no  common  roots, and  are  irreducible  in  k(p) ,the  groups  of 

f/  (x) , f^(x) , fj(x), ,and  fj (x)  will  not  have  any  common 

element  and  will  be  transitive  in  their  roots.  G,  being  the  group 
of  f(x)=0  in  k(p)  it  will  be  composed  of  these  transitive  con- 
stituents . 

If  G is  a regular  group  then  in  k(p)  all  of  the  factors  of 
f(x)  will  be  of  the  same  degree , because  a subgroup  of  a regular 
group  will  have  cycles  all  of  the  same  degree. 


If  the  group  G is  not  regular, then  all  of  the  substitutions 
do  not  contain  all  of  the  letters, and  from  a theorem*  in  group 
theory, that  the  order  of  the  sobgroup  of  a transitive  group  formed 
by  all  of  the  substitutions  which  omit  a given  letter  is  equal  to 
the  order  of  the  group  divided  by  its  degree.  Hence  in  this  case 

the  order  of  the  group  of  fj  (x)  , f^(x)  , fj(x), ,and  f5(x) 

separately  can  not  exceed  g/n, that  is, the  order  of  the  transitive 
constituents  can  not  exceed  g/n.  Likewise  if  In  k(p)  f(x)  has 
one  linear  factor, this  root  will  not  be  an  element  of  G;  ,and 
therefore  G,  will  be  of  order  g/n.  If  there  are  several  linear 
factors  in  k(p)  the  order  of  Gj  may  be  less. 

Since  Ft (V)  (i=1,2,3, ,or  t)  is  irreducible  in  k(p) , if 

the  discriminant  of  F^(V)  is  not  divisible  by  p,from  the  theory 
of  p-adic  numbers, we  have  that  the  roots  of  F^(V)=0  mod  p are 


4>  3L  jb  , / a 

MX,  - Xi/  > \i  . . % • If  S V£,=VXl  then  s\,  =SV^={SV//)  =V/J  , 

eto.  Thus  the  group  G/  is  cyclic  and  thus  transitive.  Therefore 
f(x)  has  all  linear  factors  but  one  of  degree  ,and  since  Qf 
does  not  contain  all  the  roots  its  order  is  g/n, or  a subgroup 
of  this  group  of  order  g/n. 

If  the  group  of  f(x)=0,f(x)  being  irreducible  in  R( 1 ) , is 
known, then  by  studying  the  subgroups  of  G one  can  form  some  idea 
of  the  factors  of  f(x)=0  in  k(p),and  in  many  cases  determine 
exactly  the  form  of  the  factors, as  in  the  case  when  G is  regular. 

By  extending  this  work  it  may  be  possible  to  determine  the 
exact  subgroup  for  certain  classes  of  p and  thus  determine  exact- 
ly the  factors  of  f(x)  in  k(p)  from  the  group  standpoint  of  view. 


*Miller ,Blichf eldt ,and  Dickson  'Finite  Groups’  page  32. 


